Note that this actually shows that any such piecewise linear function is in Vp for every
p ≥ 1.

Next is a fundamental approximation lemma which says that when you replace a function
in Vp with its piecewise linear approximation the p variation gets smaller.

Lemma 10.2.3Let F ∈ Vp

([0,T ])

and let P be a dissection. Then

∥∥ P ∥∥
F p,[0,T] ≤ ∥F∥p,[0,T]

Proof: Let Pε be a dissection of

[0,T]

such that

∑ | | ∥ ∥
|FP (ui+1)− F P (ui)|p > ∥FP ∥p − ε (10.1)
ui∈Pε p,[0,T ]

(10.1)

The idea is to show that, deleting some points of Pε you can assume every grid point of Pε is
also a grid point of P while preserving the above inequality. In other words, we can remove
the grid points of Pε which are not already in P and end up making the left side at least as
large. This might not be too surprising because the variation of FP is determined by the
F

(ti)

where ti∈P. Let the mesh points of Pε be denoted by

{ui}

and those of P are
denoted by

{tj}

.

Let ui+1 = u be the first point of Pε which is not in P. Then u is not equal to any of the
tj and so u ∈

(tj,tj+1)

for some j. Now let v be the very next point of P∪Pε after u. The
possibilities are

ui = tk,k ≤ j,ui+1 ∈ (tj,tj+1), v = ui+2,ui+2 ≤ tj+1 (*)

(*)

ui = tk,k ≤ j,ui+1 ∈ (tj,tj+1), v < ui+2, so v = tj+1 (**)

(**)

PICT

Consider case * In this case, you have ui+2 comes before tj+1 and so FP is linear on

[tj,v]

. Therefore, the map

| | | |
s → |FP (s)− FP (tk)|p + |F P (ui+2)− F P (s)|p

is convex and so its maximum occurs at an end point. You get for the maximum

where ui+1 is like s. Thus this sum of the two terms is dominated by either of

| |p | |p
|F P (tj)− FP (tk)| + |FP (ui+2)− F P (tj)|

or

||FP (t )− FP (t)||p + ||F P (u )− F P (t )||p
j+1 j i+2 j+1

and so if replaced with one of these, the inequality of 10.1 is preserved without using
ui+1.

Thus this eliminates ui+1 = u from Pε and preserves the inequality of 10.1. Repeat the
process until all the mesh points in Pε are in P. Now using this possibly modified Pε, each
mesh point is in P and so